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Don't understand the answer to a past paper question (aqa, june 2008) watch

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    I was looking at this question from a past paper and I don't understand how the mark scheme got to the answer.

    Q: An aqueous buffer solution which contains 0.0550 mol of a different weak acid, HA, and 0.0250 mol of NaA in 100 cm3 of solution has a pH of 4.20.

    The value of Ka for the acid HA is 2.87 × 10–5 mol dm–3 at 25 °C.

    Deduce the pH of the solution formed when 100 cm∧3 of pure water are added to 100 cm∧3 of this buffer solution.

    Mark Scheme answer: 4.20

    Surely the pH should go up because the concentration of H+ ions will have gone down. I may be wrong, if so could someone explain to me why?

    Thanks
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    The pH of a buffer is independent of the volume, take a look at his equation:

    pH = pK_a + log \frac{[A^-]}{[HA]} {derived from the equilibrium constant formula}

    the [A-]/[HA] means that the volume will always cancel so only the number of moles is relevant
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    So, in a normal solution volume does affect pH but not in a buffer. Why is that?

    Also, I don't really understand that formula. I thought pH was -log [H+].
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    (Original post by fireice7)
    So, in a normal solution volume does affect pH but not in a buffer. Why is that?
    If you have just an acid, obviously diluting that acid makes the proton concentration fall. In a buffer the amount of salt (which you don't have with a simple acid) to acid becomes important.

    Also, I don't really understand that formula. I thought pH was -log [H+].
    You don't really need to understand it :nah: that pH formula only really works for strong acids where the dissociation is nearly complete. Buffers use weak acids so you have to start from here:

    K_a = \frac{[H^+][A^-]}{[HA]}

    A bit of rearranging of this gets you the equation in my first post
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    (Original post by EierVonSatan)
    The pH of a buffer is independent of the volume, take a look at his equation:

    pH = pK_a + log \frac{[A^-]}{[HA]} {derived from the equilibrium constant formula}

    the [A-]/[HA] means that the volume will always cancel so only the number of moles is relevant
    isn't it pH = pK_a - log \frac{[A^-]}{[HA]} ?
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    (Original post by Lou Reed)
    isn't it pH = pK_a - log \frac{[A^-]}{[HA]} ?
    Nah http://en.wikipedia.org/wiki/Henders...balch_equation
 
 
 
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